Question 3.9. Can the assumption $(\heartsuit )$ be weakened to $\mathfrak b=\mathfrak s=\mathfrak c$ in Theorem 3.8?

Proof. Without loss of generality we can assume that the underlying set of X is disjoint with $\mathfrak c$ . The first countability of X implies that $|X|\leq \mathfrak c^{\omega }=\mathfrak c$ . If X is countably compact, then there is nothing to prove. Otherwise, let

$$\begin{align*}\mathcal{D}=\{A\in [X]^{\omega}: \mbox{ } A \mbox{ is closed and discrete in }X \}.\end{align*}$$

Fix any bijection $h: \mathcal {D}\cup [\mathfrak {c}]^{\omega }\rightarrow \mathfrak {c}$ such that $h(a)\geq \sup (a)$ for any $a\in [\mathfrak {c}]^{\omega }$ . It is easy to see that such a bijection exists. Next, for every $\alpha \leq \mathfrak {c}$ we shall recursively construct a topology $\tau _{\alpha }$ on $X_{\alpha }\subseteq X\cup \alpha $ . For the sake of brevity we denote the space $(X_{\alpha },\tau _{\alpha })$ by $Y_{\alpha }$ . At the end, we will show that the space $Y_{\mathfrak {c}}$ is regular first-countable countably compact and contains X as a dense subspace.

Let $X_0=X$ . Since X is first-countable and regular there exists a base $\mathcal B_0=\bigcup _{x\in X}B_{0}^x$ of the topology on X, where for each $x\in X$ , the collection $\mathcal B_{0}^x=\{U_{n,0}^x:n\in \omega \}$ is an open neighborhood base at x. With no loss of generality we can assume that $|\mathcal B_{0}|<\mathfrak c$ ; $U_{0,0}^x=X$ for each $x\in X$ ; and $\overline {U_{n+1,0}^x}\subset U_{n,0}^x$ for every $n\in \omega $ and $x\in X$ .

Assume that for each $\alpha <\xi $ regular first-countable spaces $Y_{\alpha }$ are already constructed by defining a base $\mathcal B_{\alpha }=\bigcup _{x\in X_{\alpha }}\mathcal B_{\alpha }^x$ of the the topology $\tau _{\alpha }$ , where for each $x\in X_{\alpha }$ , the collection $\mathcal B_{\alpha }^x=\{U_{n,\alpha }^x:n\in \omega \}$ is an open neighborhood base at x. Additionally assume that $X_{\alpha }\subseteq X_{\beta }$ for any $\alpha \in \beta $ and the family $\mathcal B_{\alpha }$ satisfies the following conditions:

1. (1) $|\mathcal B_{\alpha }|<\mathfrak c$ ;

2. (2) $U_{0,\alpha }^x=X_{\alpha }$ for each $x\in X_{\alpha }$ ;

3. (3) $\operatorname {cl}_{Y_{\alpha }}(U_{n+1,\alpha }^x)\subset U_{n,\alpha }^x$ for every $n\in \omega $ and $x\in X_{\alpha }$ ;

4. (4) for every $\alpha <\beta <\xi $ , $n\in \omega $ and $x\in X_{\alpha }$ we have that $U_{n,\alpha }^x=U_{n,\beta }^x\cap X_\alpha $ .

There are three cases to consider:

1. 1) $\xi =\gamma +1$ for some $\gamma \in \mathfrak {c}$ and $h^{-1}(\gamma )\cap X_{\gamma }$ is not an infinite closed discrete subset of $Y_{\gamma }$ ;

2. 2) $\xi =\gamma +1$ for some $\gamma \in \mathfrak {c}$ and $h^{-1}(\gamma )\cap X_{\gamma }$ is an infinite closed discrete subset of $Y_{\gamma }$ ;

3. 3) $\xi $ is a limit ordinal.

1) Let $X_{\xi }=X_{\gamma }$ and $\mathcal B_{\xi }=\mathcal B_{\gamma }$ .

2) Put $X_{\xi }=X_{\gamma }\cup \{\gamma \}$ . Let $h^{-1}(\gamma )=\{z_n\}_{n\in \omega }$ . For each $U_{n,\gamma }^x\in \mathcal B_{\gamma }$ consider the set

$$\begin{align*}A_{U_{n,\gamma}^x}=\{k\in\omega: z_k\in U_{n,\gamma}^x\setminus U_{n+1,\gamma}^x\}.\end{align*}$$

Since $|\mathcal B_{\gamma }|<\mathfrak c=\mathfrak s$ we get that the family $\{A_{U_{n,\gamma }^x}:U_{n,\gamma }^x\in \mathcal B_{\gamma }\}$ is not splitting, i.e., there exists a subset $A\subset \omega $ such that for each $U_{n,\gamma }^x\in \mathcal B_{\gamma }$ the set $d_{\gamma }=\{z_n:n\in A\}$ is either almost contained in $U_{n,\gamma }^x\setminus U_{n+1,\gamma }^x$ or almost disjoint with $U_{n,\gamma }^x\setminus U_{n+1,\gamma }^x$ . Since the set $\{z_n:n\in \omega \}$ is closed in $Y_{\gamma }$ for each $x\in X_{\gamma }$ there exists $n\in \omega $ such that $|U_{n,\gamma }^x\cap \{z_k:k\in \omega \}|\leq 1$ . Since $U^x_{0,\gamma }=X_{\gamma }$ for all $x\in X_{\gamma }$ , we obtain that for each $x\in X_{\gamma }$ there exists a unique $m(x)\in \omega $ such that $d_{\gamma }\subset ^* U_{m(x),\gamma }^x\setminus U_{m(x)+1,\gamma }^x$ .

For each $x\in X_\gamma $ let $\mathcal E(x)$ be the pair $(U^x_{m(x),\gamma }, U^x_{m(x)+2,\gamma })\in \mathcal B_\gamma {\times } \mathcal B_\gamma $ . Define a function $f_{\mathcal E(x)}\in \omega ^A$ as follows: if $z_n\in U^x_{m(x),\gamma }\setminus \operatorname {cl}_{Y_{\gamma }}(U^x_{m(x)+2,\gamma })$ , then

$$\begin{align*}f_{\mathcal E(x)}(n)=\min\{k: U^{z_n}_{k,\gamma}\subset U^x_{m(x),\gamma}\setminus \operatorname{cl}_{Y_{\gamma}}(U^x_{m(x)+2,\gamma})\},\end{align*}$$

and $f_{\mathcal E(x)}(n)=0$ , otherwise. Since $|\mathcal B_{\gamma }|<\mathfrak c=\mathfrak b$ there exists a function $f\in \omega ^A$ such that $f\geq ^* f_{\mathcal E(x)}$ for every $U^x_{m(x),\gamma }\in \mathcal B_{\gamma }$ . With no loss of generality we can additionally assume that $U_{f(n),\gamma }^{z_n}\cap U_{f(m),\gamma }^{z_m}=\emptyset $ for each distinct $n,m\in A$ . It is easy to see that the sequence $\{U_{f(n),\gamma }^{z_n}:n\in A\}$ is locally finite in $Y_{\gamma }$ .

Next we define an open neighborhood base at the point $\gamma \in X_{\xi }$ : Put

$$\begin{align*}U_{0,\xi}^{\gamma}=X_{\xi}\quad \mbox{ and }\quad U_{k,\xi}^{\gamma}=\bigcup_{n\in A\setminus k}U^{z_n}_{f(n)+k,\gamma}\cup\{\gamma\}\end{align*}$$

for all $k\in \mathbb N$ .

Now we are going to define an open neighborhood base $\mathcal B_{\xi }^x$ at each point $x\in X_{\gamma }$ . Let $U_{0,\xi }^x=X_{\xi }$ . For each $n\in \mathbb N$ let $U_{n,\xi }^x=U_{n, \gamma }^x$ if $n\geq m(x)+1$ and $U_{n,\xi }^x=U_{n,\gamma }^x\cup \{\gamma \}$ if $n\leq m(x)$ . It is easy to check that the family $\mathcal B_{\xi }=\{U_{n,\xi }^x: x\in X_{\xi },n\in \omega \}$ forms a base of a topology $\tau _{\xi }$ , and for each $x\in X_{\xi }$ the family $\mathcal B_{\xi }^{x}=\{U_{n,\xi }^x:n\in \omega \}$ forms an open neighborhood base at x in $Y_{\xi }$ .

At this point it is a tedious routine to check that the family $\mathcal B_{\xi }$ satisfies conditions (1)–(4).

3) Let $X_{\xi }=\bigcup _{\alpha \in \xi }X_{\alpha }$ . For each $x\in X_{\xi }$ let $\theta _x=\min \{\alpha :x\in X_{\alpha }\}$ , and for every $n\in \omega $ let us put

$$\begin{align*}U_{n,\xi}^x=\bigcup_{\theta_x\leq\alpha<\xi}U_{n,\alpha}^x.\end{align*}$$

It is quite straightforward to check that the family $\mathcal B_{\xi }=\{U_{n,\xi }^x: n\in \omega , x\in X_{\xi }\}$ forms a base of a topology $\tau _{\xi }$ , and the family $\mathcal B_{\xi }$ satisfies conditions (1), (2), and (4). In order to check the validity of condition (3) we need the following auxiliary claim.

Fix any $x\in X_{\xi }$ , $n\in \omega $ and $z\in \overline {U_{n+1,\xi }^x}$ . There exists an ordinal $\gamma \in \xi $ such that $x,z\in X_{\gamma }$ . We claim that $z\in \operatorname {cl}_{Y_{\gamma }}(U_{n+1,\gamma }^x)$ . Indeed, otherwise, there exists $m\in \omega $ such that $U^{z}_{m,\gamma }\cap U_{n+1,\gamma }^x=\emptyset $ . The above claim implies that $U^{z}_{m,\xi }\cap U_{n+1,\xi }^x=\emptyset $ , which contradicts the choice of z. Thus $z\in \operatorname {cl}_{Y_{\gamma }}(U_{n+1,\gamma }^x)\subset U_{n,\gamma }^x\subset U^x_{n,\xi }$ , which establishes condition (3).

By the construction, the space $Y_{\mathfrak {c}}$ is regular, first-countable and contains X as a dense subspace. Let A be any countable subset of $Y_{\mathfrak {c}}$ . If the set $B=A\cap \mathfrak {c}$ is infinite, then consider $h(B)\in \mathfrak {c}$ . By the construction, either B has an accumulation point in $Y_{h(B)}$ or $h(B)$ is an accumulation point of B in $Y_{h(B)+1}$ . In both cases B has an accumulation point in $Y_{\mathfrak c}$ . If $A\subset ^* X$ , then either it has an accumulation point in X, or A is closed and discrete in X. In the latter case either A has an accumulation point in $Y_{h(A)}$ or $h(A)$ is an accumulation point of A in $Y_{h(A)+1}$ . Thus $Y_{\mathfrak c}$ is countably compact.

Proof. The proof of this theorem is very similar to one of Proposition 4.1, so we give only a sketch of it. Let $\mathcal {D}$ and h be such as in the proof of Proposition 4.1.

Let $X_0=X$ . Since X is first-countable and zero-dimensional there exists a base $\mathcal B_0=\bigcup _{x\in X}\mathcal B_{0}^x$ of the topology on X consisting of clopen sets, where for each $x\in X$ , the collection $\mathcal B_{0}^x=\{U_{n,0}^x:n\in \omega \}$ is a nested open neighborhood base at x. With no loss of generality we can assume that $|\mathcal B_{0}|<\mathfrak c$ and $U_{0,0}^x=X$ for each $x\in X$ .

Assume that for each $\alpha <\xi $ Hausdorff zero-dimensional first-countable spaces $Y_{\alpha }$ are already constructed by defining a base $\mathcal B_{\alpha }=\bigcup _{x\in X_{\alpha }}\mathcal B_{\alpha }^x$ of the topology $\tau _{\alpha }$ on a set $X_{\alpha }\subseteq X\cup \alpha $ , where for each $x\in X_{\alpha }$ , the collection $\mathcal B_{\alpha }^x=\{U_{n,\alpha }^x:n\in \omega \}$ is a nested open neighborhood base at x consisting of clopen sets. Additionally assume that $X_{\alpha }\subseteq X_{\beta }$ for any $\alpha \in \beta $ and the family $\mathcal B_{\alpha }$ satisfies the following conditions:

1. (1) $|\mathcal B_{\alpha }|<\mathfrak c$ ;

2. (2) $U_{0,\alpha }^x=X_{\alpha }$ for each $x\in X_{\alpha }$ ;

3. (3) for every $\alpha <\beta <\xi $ , $n\in \omega $ and $x\in X_{\alpha }$ we have that $U_{n,\alpha }^x=U_{n,\beta }^x\cap X_\alpha $ .

There are three cases to consider:

1. 1) $\xi =\gamma +1$ for some $\gamma \in \mathfrak {c}$ and $h^{-1}(\gamma )\cap X_{\gamma }$ is not an infinite closed discrete subset of $Y_{\gamma }$ ;

2. 2) $\xi =\gamma +1$ for some $\gamma \in \mathfrak {c}$ and $h^{-1}(\gamma )\cap X_{\gamma }$ is an infinite closed discrete subset of $Y_{\gamma }$ ;

3. 3) $\xi $ is a limit ordinal.

1) Let $X_{\xi }=X_{\gamma }$ and $\mathcal B_{\xi }=\mathcal B_{\gamma }$ .

2) Put $X_{\xi }=X_{\gamma }\cup \{\gamma \}$ . Let $h^{-1}(\gamma )=\{z_n\}_{n\in \omega }$ . Similarly as in the proof of Theorem 4.1, one can find a subset $A\subset \omega $ such that for each $U_{n,\gamma }^x\in \mathcal B_{\gamma }$ the set $d_{\gamma }=\{z_n:n\in A\}$ is either almost contained in $U_{n,\gamma }^x\setminus U_{n+1,\gamma }^x$ or almost disjoint with $U_{n,\gamma }^x\setminus U_{n+1,\gamma }^x$ . Thus, for each $x\in X_{\gamma }$ there exists a unique $m(x)\in \omega $ such that $d_{\gamma }\subset ^* U_{m(x),\gamma }^x\setminus U_{m(x)+1,\gamma }^x$ .

For each $x\in X_\gamma $ let us denote by $\mathcal E(x)$ the pair $(U^x_{m(x),\gamma }, U^x_{m(x)+1,\gamma })\in \mathcal B_\gamma {\times }\mathcal B_\gamma $ . Define a function $f_{\mathcal E(x)}\in \omega ^A$ as follows: if $z_n\in U^x_{m(x),\gamma }\setminus U^x_{m(x)+1,\gamma }$ , then

$$\begin{align*}f_{\mathcal E(x)}(n)=\min\{k: U^{z_n}_{k,\gamma}\subset U^x_{m(x),\gamma}\setminus U^x_{m(x)+1,\gamma}\},\end{align*}$$

and $f_{\mathcal E(x)}(n)=0$ , otherwise. Note that $f_{\mathcal E(x)}$ is well-defined, as the set $U^x_{m(x),\gamma }\setminus U_{m(x)+1,\gamma }^x$ is open for every $x\in X_{\gamma }$ .Since $|\mathcal B_{\gamma }|<\mathfrak c=\mathfrak b$ we get that there exists a function $f\in \omega ^A$ such that $f\geq ^* f_{\mathcal E(x)}$ for every $\mathcal E(x)\in \mathcal B_{\gamma }{\times }\mathcal B_\gamma $ . We can additionally assume that $U_{f(n),\gamma }^{z_n}\cap U_{f(m),\gamma }^{z_m}=\emptyset $ for each distinct $n,m\in A$ . It is easy to see that the family $\{U_{f(n),\gamma }^{z_n}:n\in A\}$ is locally finite in $Y_{\gamma }$ .

Next we define an open neighborhood base at the point $\gamma \in X_{\xi }$ : Put

$$\begin{align*}U_{0,\xi}^{\gamma}=X_{\xi}\quad \mbox{ and }\quad U_{k,\xi}^{\gamma}=\bigcup_{n\in A\setminus k}U^{z_n}_{f(n)+k,\gamma}\cup\{\gamma\}\end{align*}$$

for all $k\in \mathbb N$ . Since the family $\{U_{f(n),\gamma }^{z_n}:n\in A\}$ is locally finite, the set $U_{k,\xi }^{\gamma }$ is clopen for each $k\in \omega $ .

Now we are going to define an open neighborhood base $\mathcal B_{\xi }^x$ at each point $x\in X_{\gamma }$ . Let $U_{0,\xi }^x=X_{\xi }$ for every $x\in X_{\gamma }$ . For each $x\in X_{\gamma }$ and $n\in \mathbb N$ let $U_{n,\xi }^x=U_{n, \gamma }^x$ if $n\geq m(x)+1$ and $U_{n,\xi }^x=U_{n,\gamma }^x\cup \{\gamma \}$ if $n\leq m(x)$ . It is easy to check that the family $\mathcal B_{\xi }=\{U_{n,\xi }^x: x\in X_{\xi },n\in \omega \}$ forms a base of a topology $\tau _{\xi }$ , and for each $x\in X_{\xi }$ the family $\mathcal B_{\xi }^{x}=\{U_{n,\xi }^x:n\in \omega \}$ forms a nested open neighborhood base at x in $Y_{\xi }$ . Moreover, the space $Y_{\xi }$ is Hausdorff and zero-dimensional.

At this point it is a tedious routine to check that the family $\mathcal B_{\xi }$ satisfies conditions (1)–(3).

3) Let $X_{\xi }=\bigcup _{\alpha \in \xi }X_{\alpha }$ . For each $x\in X_{\xi }$ let $\theta _x=\min \{\alpha :x\in X_{\alpha }\}$ . For each $x\in X_{\xi }$ and $n\in \omega $ put

$$\begin{align*}U_{n,\xi}^x=\bigcup_{\theta_x\leq\alpha<\xi}U_{n,\alpha}^x.\end{align*}$$

It is straightforward to check that the family $\mathcal B_{\xi }=\{U_{n,\xi }^x: n\in \omega , x\in X_{\xi }\}$ forms a base of a topology $\tau _{\xi }$ , satisfies conditions (1)–(3) and for each $x\in X_{\xi }$ the family $\mathcal B_{\xi }^{x}=\{U_{n,\xi }^x:n\in \omega \}$ forms an open neighborhood base at x in $Y_{\xi }$ . Hausdorffness of the space $Y_{\xi }$ follows from the next claim which can be proved similarly as Claim 4.2.

It remains to check that the space $Y_{\xi }$ is zero-dimensional which is done in the following claim.

Similarly as in the proof of Theorem 4.1 it can be checked that the space $Y_{\mathfrak {c}}$ is Hausdorff, zero-dimensional, first-countable, countably compact and contains X as a dense subspace.

Proof. Fix any Hausdorff locally compact first-countable space X of weight $< \mathfrak c=\mathfrak b$ . Without loss of generality we can assume that the underlying set of X is disjoint with $[0,1){\times }\mathfrak c$ . First countability of X implies that $|X|\leq \mathfrak c^{\omega }=\mathfrak c$ . If X is countably compact, then there is nothing to prove. Otherwise, let

$$\begin{align*}\mathcal{D}_0=\{A\in [X]^{\omega}: \mbox{ } A \mbox{ is closed and discrete in }X \}.\end{align*}$$

Fix any bijection $h:\mathfrak c{\times } \mathfrak c\rightarrow \mathfrak c$ such that $h((a,b))\geq a$ .

Next, for every $\alpha \leq \mathfrak {c}$ we shall recursively construct a topology $\tau _{\alpha }$ on $X_{\alpha }$ which will be a union of X and some pairwise disjoint family of copies of the half interval $[0,1)$ . For the sake of brevity we denote the space $(X_{\alpha },\tau _{\alpha })$ by $Y_{\alpha }$ . At the end, we will show that the space $Y_{\mathfrak {c}}$ has the desired properties.

Let $X_0=X$ and fix any bijection $g_0: \mathcal {D}_0\rightarrow \{0\}{\times }\mathfrak {c}$ . Note that by the definition of h, $h^{-1}(0)\in \{0\}{\times }\mathfrak c$ , so $g_0^{-1}(h^{-1}(0))$ is well-defined (and belongs to $\in \mathcal D_0$ ). By Corollary 4.8, the space X has property D. It follows that the closed discrete set $\{z_n:n\in \omega \}=g_0^{-1}(h^{-1}(0))$ can be expanded to a discrete family $\{U_n:n\in \omega \}$ of open sets such that $z_n\in U_n$ . Since X is locally compact, we lose no generality assuming that $\operatorname {cl}_{X}(U_n)$ is compact for every $n\in \omega $ . For each $n\in \omega $ fix a continuous function $f_n$ such that $f(z_n)=0$ and $f(X\setminus U_n)=1$ . Let $X_1=X_0\cup ([0,1){\times }\{1\})$ , and $\tau _1$ be a topology on $X_1$ which satisfies the following conditions:

1. (1) $X_0$ is an open subspace of $X_1$ ;

2. (2) the sets

   $$\begin{align*}W_n(0,1)=[0,1/n){\times}\{1\}\cup \bigcup_{m\geq n}f_m^{-1}([0,1/n)),\end{align*}$$
   $n\in \mathbb N$ form an open neighborhood base at the point $(0,1)\in X_1$ ;

3. (3) For $0<x<1$ the sets

   $$\begin{align*}W_n(x,1)=(x-1/n,x+1/n){\times}\{1\}\cup\bigcup_{m\geq n}f_m^{-1}((x-1/n,x+1/n)),\end{align*}$$
   where $n\in \mathbb N$ satisfies $(x-1/n,x+1/n)\subseteq (0,1)$ , form an open neighborhood base at the point $(x,1)\in X_1$ .

It is easy to check that $Y_1$ is a first-countable Hausdorff space of weight $<\mathfrak b$ . Observe that $(0,1)$ is an accumulation point of the set $\{z_n:n\in \omega \}=g_0^{-1}(h^{-1}(0))$ in $Y_1$ . Clearly, $Y_1$ is locally compact at each point $x\in X_0$ . Let us show that $Y_1$ is locally compact at $(0,1)$ . We claim that the set $\operatorname {cl}_{Y_1}(W_2(0,1))$ is compact. Since the family $\{U_n:n\in \omega \}$ is discrete and the sets $U_n$ , $n\in \omega $ have compact closure in X, it is easy to see that the set

$$\begin{align*}\operatorname{cl}_{Y_1}(W_2(0,1))\subseteq [0,1/2]{\times}\{1\}\cup\bigcup_{n\in\omega\setminus\{0,1\}}\operatorname{cl}_X(U_n)\end{align*}$$

is $\sigma $ -compact. At this point it is enough to show that $\operatorname {cl}_{Y_1}(W_2(0,1))$ is countably compact. Fix any infinite subset $A=\{a_n:n\in \omega \}\subset \operatorname {cl}_{Y_1}(W_2(0,1))$ . If for some $n\in \omega $ the set $A\cap \operatorname {cl}_X(U_n)$ is infinite or the set $A\cap ([0,1/2){\times }\{1\})$ is infinite, then A has an accumulation point in $Y_1$ . So, let us assume that all the aforementioned intersections are finite. Shrinking the set A if necessary, we can assume that $A\subset X$ and $|A\cap U_n|\leq 1$ for all $n\in \omega $ . For each $n\in \omega $ let $m(n)$ be such that $a_n\in U_{m(n)}$ . Put $y_n=f_{m(n)}(a_n)$ . Since $\{y_n:n\in \omega \}\subset [0,1/2)$ and the interval $[0,1/2]$ is compact there exists $z\in [0,1/2]$ and a subsequence $\{y_{n_k}:k\in \omega \}$ which converges to z. It is easy to see that $(z,1)$ is an accumulation point of the set A in $Y_1$ . Hence the $\sigma $ -compact set $\operatorname {cl}_{Y_1}(W_2(0,1))$ is countably compact and thus compact. Similarly one can check that $Y_1$ is locally compact at $(x,1)$ for each $0<x<1$ . Let

$$\begin{align*}\mathcal{D}_1=\{A\in [X_1]^{\omega}: \mbox{ } A \mbox{ is closed and discrete in }Y_1 \}.\end{align*}$$

Fix any bijection $g_1: \mathcal {D}_1\rightarrow \{1\}{\times }\mathfrak {c}$ .

Assume that locally compact Hausdorff spaces $Y_\alpha =(X_{\alpha },\tau _{\alpha })$ with weight $<\mathfrak b$ are constructed for all $\alpha \in \xi $ such that $Y_\alpha $ is an open dense subspace of $Y_{\eta }$ for all $\alpha \in \eta \in \xi $ . Additionally assume that if $X_{\alpha +1}\setminus X_{\alpha }\neq \emptyset $ , then $X_{\alpha +1}\setminus X_{\alpha }=[0,1){\times }\{\alpha +1\}$ and for each $\alpha \in \xi $ it is defined a bijection $g_{\alpha }: \mathcal {D}_{\alpha }\rightarrow \{\alpha \}{\times }\mathfrak {c}$ , where

$$\begin{align*}\mathcal{D}_{\alpha}=\{A\in [X_{\alpha}]^{\omega}: \mbox{ } A \mbox{ is closed and discrete in }Y_{\alpha} \}.\end{align*}$$

If $\xi $ is a limit ordinal, then put $X_{\xi }=\bigcup _{\alpha \in \xi }X_{\alpha }$ and $\tau _{\xi }=\bigcup _{\alpha \in \xi }\tau _{\alpha }$ . Clearly, $Y_\xi $ is first-countable, Hausdorff, locally compact with weight less than $\mathfrak b$ . Fix any bijection $g_{\xi }: \mathcal D_\xi \rightarrow \{\xi \}{\times }\mathfrak c$ .

Assume that $\xi =\gamma +1$ for some $\gamma \in \mathfrak {c}$ . Consider the pair $h^{-1}(\gamma )=(a,b)\in \mathfrak c{\times }\mathfrak c$ . If the set $g_a^{-1}(h^{-1}(\gamma ))\in \mathcal D_a$ already has an accumulation point in $Y_{\gamma }$ , then put $Y_{\xi }=Y_{\gamma }$ . If the set $g_a^{-1}(h^{-1}(\gamma ))$ is closed and discrete in $Y_{\gamma }$ , then we construct a Hausdorff locally compact first-countable space $Y_{\xi }$ of weight $<\mathfrak b$ similarly as we constructed $Y_1$ . In particular, $X_{\xi }=X_{\gamma }\cup [0,1){\times }\{\xi \}$ , and $(0,\xi )$ is an accumulation point of the set $g_a^{-1}(h^{-1}(\gamma ))$ .

Obviously, $Y_{\mathfrak c}$ is a Hausdorff locally compact first-countable space. To derive a contradiction, assume that $Y_{\mathfrak c}$ is not countably compact. Then $Y_{\mathfrak c}$ contains an infinite countable discrete closed subset A. Since the cardinal $\mathfrak b=\mathfrak c$ is regular, there exists an ordinal $\xi \in \mathfrak c$ such that $A\subset X_{\xi }$ . Obviously, A is an infinite closed discrete subset of $Y_{\beta }$ for each $\beta \geq \xi $ . Then $g_{\xi }(A)=(\xi ,\delta )$ for some $\delta \in \mathfrak c$ . Let $\mu \in \mathfrak c$ be such that $\xi \leq \mu $ and $h((\xi ,\delta ))=\mu $ . By the construction, the point $(0,\mu +1)\in X_{\mu +1}$ is an accumulation point of A in $Y_{\mathfrak c}$ , which implies the desired contradiction. Hence the space $Y_{\mathfrak c}$ is countably compact.

Proof. Fix a base $\mathcal B$ of X of size $<\min \{\mathfrak {s},\mathfrak {b}\}$ consisting of clopen sets. We lose no generality assuming that $\mathcal B=\{U_n^x:x\in X, n\in \omega \}$ , where for every $x\in X$ the family $\{U_n^x:n\in \omega \}$ forms a nested open neighborhood base at x. By Proposition 4.10 it is enough to check that for every $T\in \mathbf {D}(\mathcal B)$ there exists $S\in \mathbf {E}(\mathcal B)$ such that S refines T. Consider an arbitrary sequence $T=\{T_n:n\in \omega \}\in \mathbf {D}(\mathcal B)$ and for each $n\in \omega $ fix a point $x_n\in T_n$ . To each $B\in \mathcal B$ we assign a function $f_B\in \omega ^\omega $ defined as follows: if $x_n\in B$ , then put $f_B(n)=\min \{k\in \omega : U_k^{x_n}\subset B\cap T_n\}$ ; otherwise put $f_B(n)=\min \{k\in \omega : U_k^{x_n}\subset T_n$ and $U_k^{x_n}\cap B=\emptyset \}$ . Since $|\mathcal B|<\mathfrak b$ , there exists a function $f\in \omega ^\omega $ such that $f\geq ^* f_B$ for every $B\in \mathcal B$ . Consider the refinement $\{U^{x_n}_{f(n)}:n\in \omega \}$ of T. Clearly, for each $B\in \mathcal B$ there exists $k\in \omega $ such that for every $n\geq k$ either $U^{x_n}_{f(n)}\cap B=\emptyset $ or $U^{x_n}_{f(n)}\subset B$ . So, to each $B\in \mathcal B$ we can assign a set $C_B=\{n\in \omega : U^{x_n}_{f(n)}\subset B\}\subset \omega $ . Since $|\mathcal B|<\mathfrak s$ we get that the family $\{C_B:B\in \mathcal B\}$ is not splitting. Hence there exists an infinite subset A of $\omega $ possessing the following property: for every $B\in \mathcal B$ there exists $k\in \omega $ such that either $U^{x_n}_{f(n)}\subset B$ for each $n\in A\setminus k$ or $U^{x_n}_{f(n)}\cap B=\emptyset $ for each $n\in A\setminus k$ . Hence the sequence $S=\{U^{x_n}_{f(n)}:n\in A\}$ is a $\mathcal B$ -exact refinement of T.

Proof. Fix an unbounded subset $\{b_{\alpha }:\alpha \in \mathfrak b\}\subset \omega ^\omega $ which satisfies the following conditions:

1. (1) the function $b_{\alpha }$ is increasing for every $\alpha \in \mathfrak b$ ;

2. (2) $b_\alpha <^* b_{\beta }$ whenever $\alpha \in \beta $ .

For $i\in \omega $ set $v_i=\{(i,n):n\in \omega \}$ . Let . Condition (2) implies that $\mathcal A$ is an almost disjoint family of subsets of $\omega {\times }\omega $ . To derive a contradiction, assume that there exists a first-countable Hausdorff countably compact space X which contains the Mrowka space $\psi (\mathcal A)$ as a subspace. Find an infinite subset C of $\omega $ such that the sequence $\{v_{n}:n\in C\}$ converges to a point $x\in X$ . Fix an open neighborhood base $\{U_n:n\in \omega \}$ at x. Without loss of generality we can assume that $\{v_k:k\in C\setminus n\}\subset U_n$ for all $n\in \omega $ . Observe that for each $n\in \omega $ there exists a function $f_n\in \omega ^C$ such that $\{(k,m): k\in C\setminus n, m\geq f_n(k)\}\subset U_n$ . Since the functions $b_{\alpha }$ , $\alpha \in \mathfrak b$ are increasing, the family $\{b_\alpha {\restriction }_C :\alpha \in \mathfrak b\}$ is unbounded. Then for each $n\in \omega $ there exists $\alpha _n\in \mathfrak b$ such that $f_n\not \geq ^* b_{\alpha _n}{\restriction }_C$ . It follows that for each $n\in \omega $ and $\xi \geq \alpha _n$ ,

$$\begin{align*}|\{(k,m): k\in C\setminus n, m\geq f_n(k)\}\cap b_{\xi}|=\omega.\end{align*}$$

Thus for $\mu = \sup \{\alpha _n:n\in \omega \}$ the set $b_{\mu }\cap \{(k,m): k\in C\setminus n, m\geq f_n(k)\}$ is infinite for all $n\in \omega $ . Hence $b_{\mu }\in \overline {U_n}$ for all $n\in \omega $ , which contradicts the Hausdorffness of X.

Proof. By $\mathbb Q$ we denote the set of rational numbers from the interval $[0,1]$ . Enumerate the set of all subsets of $\omega $ as $\{A_x:x\in [0,1]\}$ , and for each $x\in [0,1]$ and $m\in \mathbb N$ choose $q_x(m)\in \mathbb Q$ such that $0<|x-q_x(m)|<1/m$ and put $Q_x=\{q_x(m):m\in \mathbb N\}$ . Let X be the set $[0,1]{\times }\{0\}\cup \mathbb Q{\times }\mathbb {\mathbb N}$ endowed with a topology $\tau $ satisfying the following conditions:

• • $\mathbb Q{\times }\mathbb {\mathbb N}$ is the set of isolated points;

• • if $x\in \mathbb Q$ , then basic neighborhoods of a point $(x,0)\in [0,1]{\times }\{0\}$ have the form

  $$\begin{align*}B_{m}(x)=\{(a,b)\in X: |x-a|<1/m\}\setminus (Q_x{\times}A_x\cup \{x\}{\times}\mathbb N) \qquad \mbox{for} \quad m\in\mathbb N;\end{align*}$$

• • if $x\in [0,1]\setminus \mathbb Q$ , then basic neighborhoods of a point $(x,0)\in [0,1]{\times }\{0\}$ have the form

  $$\begin{align*}B_{m}(x)=\{(a,b)\in X: |x-a|<1/m\}\setminus (Q_x{\times}A_x) \qquad \mbox{for} \quad m\in\mathbb N.\end{align*}$$

Clearly, the spaceFootnote ¹ X is regular, separable and first-countable.

Fix an arbitrary splitting family $\mathcal S\subset [\omega ]^\omega $ of cardinality $\mathfrak s$ and consider the subspace

$$\begin{align*}Y=\{x\in[0,1]: A_x\in \mathcal S\}{\times}\{0\}\cup\mathbb Q{\times}\mathbb N\end{align*}$$

of X. It is clear that Y is regular and first-countable. Taking into account that the subspace $([0,1]{\times }\{0\})\cap Y$ of Y is second-countable, we deduce that Y is separable, Lindelöf and subsequently normal.

Let us show that Y doesn’t embed densely into a first-countable regular feebly compact space. Assuming the contrary, fix any first-countable regular feebly compact space Z which contains Y as a dense subspace. The following claim is crucial.

Fix any enumeration $\mathbb Q=\{q_n:n\in \omega \}$ . Since the set is open and discrete in a feebly compact space Z, there exists an accumulation point $z_0$ of $Q_0$ . Since Z is first-countable, there exists $A_0\subseteq \mathbb N$ such that the sequence $\{(q_0,n):n\in A_0\}$ converges to $z_0$ . Assume that for each $i\leq n$ we constructed a subset $A_i$ of $\mathbb N$ satisfying:

1. (1) $A_i\subset A_j$ for every $i\leq j$ ;

2. (2) the sequence $\{(q_i,n):n\in A_i\}$ converges to some $z_i\in Z$ .

Consider the discrete open set . Since Z is first-countable and feebly compact, there exists an infinite subset $A_{n+1}\subset A_n$ such that the sequence $\{(q_{n+1},n):n\in A_{n+1}\}$ converges to some point $z_{n+1}\in Z$ .

After completing the induction we obtain a decreasing chain $\{A_i:i\in \omega \}$ of infinite subsets of $\mathbb N$ and a sequence of points $\{z_i:i\in \omega \}$ such that $z_i$ is the limit of the sequence $\{(q_i,n):n\in A_i\}$ . Let A be any pseudointersection of the family $\{A_i:i\in \omega \}$ , i.e., $A\subseteq ^* A_i$ for all $i\in \omega $ . It is easy to see, that the sequence $\{(q_i,n):n\in A\}$ converges to $z_i$ for each $i\in \omega $ . But this contradicts Claim 5.3.

Proof. Let $B\setminus A=\{b_n:n\in \omega \}$ and $A\cup B=\bigcap _{n\in \omega }U_n$ , where $U_n$ is open for each $n\in \omega $ . Then the set is open for each $n\in \omega $ , and $\bigcap _{n\in \omega } V_n=A$ , i.e., A is $G_{\delta }$ .

Proof. Let A be a countable infinite subset of X which is not $G_\delta $ . Seeking a contradiction, assume that $\operatorname {PR}(X)$ embeds densely into a regular first-countable feebly compact space Z. Fix any countable dense subset $A'$ of X. Lemma 5.4 implies that $Q=A\cup A'$ is not $G_{\delta }$ . Fix an increasing sequence $\{F^{\emptyset }_k:k\in \omega \}$ of finite subsets of Q such that $\bigcup _{k\in \omega }F^{\emptyset }_k=Q$ . Let . Consider the family $\{\operatorname {cl}_Z[F_k^{\emptyset }, X]: k\in \omega \}$ . Taking into account that $\operatorname {PR}(X)$ is dense in Z and $[F_k^{\emptyset }, B_{\emptyset }]$ is clopen in $\operatorname {PR}(X)$ , it is easy to check that $[F_k^{\emptyset }, B_{\emptyset }]\subset \operatorname {int}_Z(\operatorname {cl}_Z[F_k^{\emptyset }, B_{\emptyset }])$ for all $k\in \omega $ . So, the family $\{\operatorname {int}_Z(\operatorname {cl}_Z[F_k^{\emptyset }, B_{\emptyset }]):k\in \omega \}$ consists of open subsets of Z and is centered. By feeble compactness of Z, there exists

$$\begin{align*}z_\emptyset\in \bigcap_{k\in\omega}\operatorname{cl}_Z(\operatorname{int}_Z(\operatorname{cl}_Z[F_k^{\emptyset}, B_{\emptyset}]))=\bigcap_{k\in\omega}\operatorname{cl}_Z[F_k^{\emptyset}, B_{\emptyset}].\end{align*}$$

Fix a countable open neighborhood base $\{O^{\emptyset }_k:k\in \omega \}$ at $z_{\emptyset }$ . Since $\operatorname {PR}(X)$ is dense in Z, for each $k\in \omega $ there exists a basic clopen subset $[G_k^{\emptyset }, W_k^{\emptyset }]$ of $\operatorname {PR}(X)$ such that $[G_k^{\emptyset }, W_k^{\emptyset }]\subseteq O_k^{\emptyset }\cap [F_k^{\emptyset }, B_{\emptyset }]$ . Clearly, $F_{k}^{\emptyset }\subseteq G_{k}^{\emptyset }$ and the sequence $\{G_{k}^{\emptyset }:k\in \omega \}\subset \operatorname {PR}(X)$ converges to $z_{\emptyset }$ .

Suppose that for some $i\geq 0$ we accomplished the following things:

1. (1) constructed a subtree $\mathbb T_i$ of $\omega ^{\leq i}$ ;

2. (2) constructed a family $\{B_s:s\in \mathbb T_i\}$ of clopen subsets of X satisfying the following conditions:

   1. (2.1) for every $s\in \mathbb T_i\cap \omega ^{<i}$ , , where ;

   2. (2.2) $B_{s\hat {~}n}\cap B_{s\hat {~}m}=\emptyset $ for every $s\in \mathbb T_i\cap \omega ^{<i}$ and distinct $n,m\in \beta _s$ ;

   3. (2.3) for each $s\in \mathbb T_i$ the diameter of $B_s$ doesn’t exceed $1/|s|$ .

3. (3) $\forall s\in \mathbb T_i$ fixed an increasing sequence $\{F_k^s:k\in \omega \}$ of finite subsets of such that $\bigcup _{k\in \omega }F_k^s= Q_s$ ;

4. (4) $\forall s\in \mathbb T_i$ fixed a point $z_s\in \bigcap _{k\in \omega }\operatorname {cl}_Z[F_k^s,B_s]$ and a nested open neighborhood base $\{O_k^s:k\in \omega \}$ at $z_s$ ;

5. (5) $\forall s\in \mathbb T_i$ , $\forall k\in \omega $ fixed a subset $[G^s_k,W_k^s]\subset O_k^s\cap [F_k^s,B_s]$ such that the following condition holds:

   1. (5.1) $\forall s\in \omega ^{<i}\cap \mathbb T_i$ , $\forall n\in \beta _s$ , $\exists k(s,n)\geq |s|+1$ such that $B_{s\hat {~}n}\subseteq \bigcap _{j\leq |s|}\times W_{k(s,n)}^{s{\restriction }j}$ .

Fix any $s\in \mathbb T_i\cap \omega ^{i}$ . Since $B_s$ is open in X and Q is dense, the set is not empty. Note that since X could contain isolated points, it is possible that $|Q_s|<\omega $ . So, let us fix an enumeration $Q_s=\{q_n^s:n\in \alpha _s\}$ , where $\alpha _s\in \omega +1$ . Observe that $q^s_0\in Q\cap B_{s{\restriction }j}$ for all $j\leq |s|$ . Since $\{F_k^{s{\restriction }j}: k\in \omega \}$ is an increasing sequence of finite subsets of $Q\cap B_{s{\restriction }j}$ , there exists $n_j\in \omega $ such that $q^s_0\in F_{n}^{s{\restriction }j}$ for all $n\geq n_j$ . Put $k(s,0)=\max \{n_j:j\leq |s|\}+|s|+1$ and note that the open set $\bigcap _{j\leq |s|} W_{k(s,0)}^{s{\restriction }j}$ contains $q^s_0$ . So fix an arbitrary clopen set $B_{s\hat {~}0}\subseteq \bigcap _{j\leq |s|} W_{k(s,0)}^{s{\restriction }j}$ which contains $q_0^s$ and has diameter $\leq 1/(i+1)$ . By condition (5), we get that $W_{k(s,0)}^s\subseteq B_{s}$ which implies that $B_{s\hat {~}0}\subseteq B_s$ . If $Q_s\subset B_{s\hat {~}0}$ , we put $\beta _s=1$ and move to another $s\in \mathbb T_i\cap \omega ^{i}$ . If $Q_s\setminus B_{s\hat {~}0}\neq \emptyset $ , find minimal $m\in \omega $ such that $q_m^s\notin B_{s\hat {~}0}$ . Similarly as above there exists $k(s,1)\geq |s|+1$ such that the open set $\bigcap _{j\leq |s|} W_{k(s,1)}^{s{\restriction }j}$ contains $q^s_m$ . Since the set $B_{s\hat {~}0}$ is clopen and doesn’t contain $q_m^s$ , we can find a clopen neighborhood $B_{s\hat {~}1}$ of $q_m^s$ of diameter $\leq 1/(i+1)$ which is contained in $(B_s\setminus B_{s\hat {~}0})\cap \bigcap _{j\leq |s|} W_{k(s,1)}^{s{\restriction }j}$ . If $Q_s\subset B_{s\hat {~}0}\cup B_{s\hat {~}1}$ , we put $\beta _s=2$ and move to another $s\in \mathbb T_i\cap \omega ^i$ . Otherwise, we can find minimal $m\in \omega $ such that $q_m^s\in B_s\setminus (B_{s\hat {~}0}\cup B_{s\hat {~}1})$ and repeat the previous steps. Proceeding this way we shall construct a family $\{B_{s\hat {~}k}:k\in \beta _s\in \omega +1\}$ of clopen subsets of X which satisfies conditions (2.1), (2.2), (2.3), and (5.1). Note that $\beta _s\leq \alpha _s$ can be finite. Let $\mathbb T^s_{i+1}=\mathbb T_i\cup \{s\hat {~}k:k\in \beta _s\}$ .

After making the above construction for all $s\in \mathbb T_i\cap \omega ^{i}$ , we obtain a tree $\mathbb T_{i+1}=\bigcup _{s\in \mathbb T_i\cap \omega ^{i}}\mathbb T_{i+1}^s$ and a family $\{B_{s\hat {~}k}:k\in \beta _s \mbox { and } s\in \mathbb T_i\cap \omega ^{i}\}$ of clopen subsets of X which satisfies conditions (2.1), (2.2), (2.3), and (5.1).

For each $s\in \mathbb T_{i+1}\cap \omega ^{i+1}$ fix an increasing sequence $\{F_k^s:k\in \omega \}$ of finite subsets of $Q_s$ such that $\bigcup _{k\in \omega }F_k^s=Q_s$ . Analogous arguments to those we used to find $z_\emptyset $ imply that for each $s\in \mathbb T_{i+1}\cap \omega ^{i+1}$ there exists a point $z_s\in \bigcap _{k\in \omega }\operatorname {cl}_Z[F_k^s,B_s]$ . By the first-countability of Z, we can fix a nested open neighborhood base $\{O_k^s:k\in \omega \}$ at $z_s$ . For each $s\in \mathbb T_{i+1}\cap \omega ^{i+1}$ and $k\in \omega $ fix a subset $[G^s_k,W_k^s]\subset O_k^s\cap [F_k^s,B_s]$ . This finishes step $i+1$ of the induction.

Upon completing the induction we obtain a tree $\mathbb T=\bigcup _{i\in \omega }\mathbb T_i$ , and a family $\{B_s:s\in \mathbb T\}$ of clopen subsets of X which satisfies conditions (2.1), (2.2), (2.3), and (5.1). Also, for each $s\in \mathbb T$ we found an element $z_s\in Z$ which satisfies condition (4) and constructed the families $\{W_k^s: k\in \omega , s\in \mathbb T\}$ (consisting of open subsets of X) and $\{G_k^s: k\in \omega , s\in \mathbb T\}$ (consisting of finite subsets of X), which satisfy conditions (5) and (5.1).

Let

$$\begin{align*}P=\bigcap_{i\in\omega}\bigcup_{s\in\omega^i\cap \mathbb T}B_s.\end{align*}$$

By the construction, P is a $G_{\delta }$ -set containing Q. It follows that $|P|>\omega $ . Indeed, otherwise we would have that $Q=P\cap \bigcap _{x\in P\setminus Q}(X\setminus \{x\})$ is a $G_\delta $ subset of X, which is not the case. Thus, we can fix

(1) $$ \begin{align} p_{\infty}\in P\setminus \bigcup_{s\in \mathbb T}\bigcup_{k\in\omega}G_k^s. \end{align} $$

It is easy to see that there exists $t\in \omega ^\omega $ such that $p_{\infty }\in B_{t{\restriction }i}$ for any $i\in \omega $ . Note that since the diameter of $B_{t{\restriction }i}$ doesn’t exceed $1/i$ , we get that the family $\{B_{t{\restriction }i}:i\in \omega \}$ forms an open neighborhood base at $p_{\infty }$ in X. It follows that the family $\{[\{p_{\infty }\},B_{t{\restriction }i}]:i\in \omega \}$ forms an open neighborhood base of $\{p_{\infty }\}$ in $\operatorname {PR}(X)$ . Put $R=\operatorname {int}_Z(\operatorname {cl}_Z([\{p_{\infty }\}, X]))$ . One can check that $R\cap \operatorname {PR}(X)=[\{p_{\infty }\}, X]$ . By the regularity of Z, there exists an open in Z neighborhood W of $\{p_{\infty }\}$ such that $\operatorname {cl}_Z(W)\subseteq R$ . Thus, there exists $i\in \omega $ such that $\operatorname {cl}_Z([\{p_{\infty }\}, B_{t{\restriction }i}])\subseteq R$ .

Since $\operatorname {cl}_Z([\{p_{\infty }\}, B_{t{\restriction }i}])\subseteq R$ , Claim 5.6 implies that $z_{t{\restriction }i}\in R$ . Condition (5) implies that for each $s\in \mathbb T$ the sequence $\{G_k^s:k\in \omega \}\subseteq \operatorname {PR}(X)$ converges to $z_s$ . Since the set R is an open neighborhood of $z_{t{\restriction }i}$ , there exists $k\in \omega $ such that $G_k^{t{\restriction }i}\in R\cap \operatorname {PR}(X)=[\{p_{\infty }\}, X]$ , which is impossible, as $p_\infty \notin \bigcup _{s\in \mathbb T}\bigcup _{k\in \omega }G_k^s$ (see Equation 1).